Two point charges of equal magnitude but opposite sign are placed at a certain distance. The force between them is $F$. If $75\%$ of the charge from one is transferred to the other,what will be the new force between them?

An oil drop of $12$ excess electrons is held stationary under a constant electric field of $2.55 \times 10^{4} \; N \, C^{-1}$ (Millikan's oil drop experiment). The density of the oil is $1.26 \; g \, cm^{-3}$. Estimate the radius of the drop. $(g = 9.81 \; m \, s^{-2}; e = 1.60 \times 10^{-19} \; C)$

Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another point charge $-20 \mu C$ is placed at the origin and given a small displacement $x$ $(x \ll a)$ along $X$-axis. The force acting on the point charge is ($x$ and $a$ are in metres, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

Three charges,each $+Q$,are placed at the vertices of an equilateral triangle of side $a$. What is the net electrostatic force on any one charge? $\left( k = \frac{1}{4\pi \varepsilon _0} \right)$

Three charges $q$,$Q$,and $4q$ are placed in a straight line of length $l$ at points distant $0$,$\frac{l}{2}$,and $l$ respectively from one end. In order to make the net force on $q$ zero,the charge $Q$ must be equal to

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium,assuming that the angle made by each thread is small,the separation $x$ between the balls is proportional to